4-force, 4-momentum, energy and mass relations.

AI Thread Summary
The discussion focuses on the concepts of 4-force, 4-momentum, and their relationships in Special Relativity. It begins with defining 4-force as the proper-time variation of 4-momentum and seeks to justify Einstein's mass-energy equivalence, E = mc², while deriving the relation E² = (mc²)² + (p²c²). The participant explores the implications of 3-force and 3-acceleration not being parallel, noting that the scaling factors for these components differ due to the gamma factor. The conversation also touches on the interpretation of force in Special Relativity compared to Newtonian mechanics, emphasizing the need for a different approach to force calculations. Overall, the thread highlights the complexities of relativistic dynamics and the importance of understanding these fundamental relationships.
heitor
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This is an exercise of Special Relativity the professor asked last week.
Sorry for the long post, I hope you don't get bored reading it, also, this is my first post here :shy:

Homework Statement


Defining the 4-force that acts on a particle as the proper-time variation of the 4-momentum F^\mu := \frac{d P^\mu}{d \tau}.
  1. Justify Einsteins relation between mass and energy: E = mc^2
  2. Show, using the \eta _{\mu\nu}P^\mu P^\nu, that E^2 = (mc^2)^2 + (\vec{p}^2c^2), where \vec{p} is the 3-momentum.
  3. Show that in SR the 3-force and 3-acceleration is not always parallel to each other.

Homework Equations


I'm using the convention: x^0 = ct and (- + + +) for the metric tensor.

4-velocity: u^\mu = \gamma_{(v)} (c, \vec{v}), with \vec{v} the 3-velocity in lab frame.

And some results I got, they are somewhere in Wikipedia and some books also.

gamma: \frac{d}{d \tau} = \frac{d t}{d \tau} \frac{d}{d t} = \gamma_{(v)} \frac{d}{d t}

I'll omit (v) in gamma for brevity.

derivative of gamma: \dot{\gamma} = \frac{d \gamma}{dt} = \gamma ^3 \frac{\vec{v} \cdot \vec{a}}{c^2}

acceleration: a^\mu = \left ( \gamma ^4 \frac{\vec{v} \cdot \vec{a}}{c}, \gamma^4 \vec{a}_\parallel + \gamma ^2 \vec{a}_\perp \right )
where \vec{a}_\parallel = (\vec{a} \cdot \vec{v}) \vec{v}/v^2 is the parallel component of the 3-acceleration to the 3-velocity and \vec{a}_\perp = \vec{a} - \vec{a}_\parallel is the perpendicular one.​


3. The Attempt at a Solution for question 1 and 2

I think I got the results, but i have some doubts.

F^\mu = \frac{d P^\mu}{d \tau} = \left(\frac{d}{d \tau}(m \gamma c), \frac{d}{d \tau}(m \gamma \vec{v}) \right )
I'll call this 'result' as (EQ1).

Using the gamma relation in the relevant equations:
F^\mu = \gamma \left ( m c \dot{\gamma}, \frac{d\vec{p}}{dt} \right ) = \left ( m c \gamma ^4 \frac{\vec{v} \cdot \vec{a}}{c^2}, \gamma \vec{f}_R \right )
I'll call this 'result' as (EQ2). Where \vec{f}_R is the 3-force. We know that F^\mu is mass times acceleration:

F^\mu = m a^\mu = m \left ( \gamma ^4 \frac{\vec{v} \cdot \vec{a}}{c}, \gamma^4 \vec{a}_\parallel + \gamma ^2 \vec{a}_\perp \right )
This is (EQ3).

Comparing (EQ2) and (EQ3) we get:
\vec{f}_R = m \left ( \gamma^3 \vec{a}_\parallel + \gamma \vec{a}_\perp \right )

scalar product with \vec{v} gives a variation in energy:
\vec{f}_R \cdot \vec{v} = m \gamma^3 \vec{a} \cdot \vec{v} = \frac{dE}{d \tau}
This is (EQ4).
Here is the derivative with respect to the proper-time, right? I'm not sure about this...​

Putting (EQ4) inside (EQ3) in the first component and re-using the second component of (EQ2):
F^\mu = \gamma \left (\frac{\vec{f}_R \cdot \vec{v}}{c}, \vec{f}_R \right ) = \left ( \frac{1}{c} \frac{d E}{d\tau}, \frac{d\vec{p}}{d\tau} \right )
This is (EQ5).

Comparing (EQ5) and (EQ1) yelds:
\frac{dE}{d\tau} = \frac{d}{d\tau} (m \gamma c^2)

So:

E = m \gamma c^2 +\ constant
This is (EQ6).

This is almost what the question 1 asks, but what is this constant? What is the meaning of E? Is it kinect + 'rest energy'?​

Question 2 I got:

\eta _{\mu\nu}P^\mu P^\nu = m^2 \gamma ^2 (-c^2 + \vec{v}^2) = -m^2 c^2
(EQ7)

From (EQ6) we can write the 4-momentum as:
P^\mu = \left ( \frac{E}{c}, \vec{p} \right )
(EQ8)

so:
\eta _{\mu\nu}P^\mu P^\nu = - \frac{E^2}{c^2} + \vec{p}^2 = -m^2 c^2

Wich is the answer to question 2 if the constant of (EQ6) is zero...



4. The attempt at a solution for question 3

If we do the same trick used to split the acceleration in parallel and perpendicular parts:

\vec{f}_R := \vec{f}_{R, \parallel} + \vec{f}_{R, \perp}
We build the parallel one:

\vec{f}_{R, \parallel} = \frac{(\vec{f}_R \cdot \vec{a}) \vec{a}}{\vec{a}^2}
and the perpendicular one:

\vec{f}_{R, \perp} = \vec{f}_R - \vec{f}_{R, \parallel} = \vec{f}_R - \frac{(\vec{f}_R \cdot \vec{a}) \vec{a}}{\vec{a}^2}

But I was unable to prove \vec{f}_{R, \perp} \neq \vec{0}.

Do you have any hint? All that I got was some previous equations.


If I was not clear in some statement, please tell me.

(is there a way to put a 'name' in some equations to be displayed on the right side of it? Like latex documents?)
(Sorry for my bad English...)

Thanks in advance,
Heitor.
 
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From the expression \vec{f}_R = m \left ( \gamma^3 \vec{a}_\parallel + \gamma \vec{a}_\perp \right ) we see that the force is not parallel to the acceleration because \gamma^3 is different from \gamma so the parallel and perpendicular components scale differently.
 
By the way, I think that instead of taking equations from books and wikipedia, you ought to prove them.
 
dauto said:
From the expression \vec{f}_R = m \left ( \gamma^3 \vec{a}_\parallel + \gamma \vec{a}_\perp \right ) we see that the force is not parallel to the acceleration because \gamma^3 is different from \gamma so the parallel and perpendicular components scale differently.

Is there any interpretation for this? A 3-force in Special Relativity does not have the same meaning as in Newtonian Mechanics?


I did proved them, it was the previous question in the list.
 
The 3-force is derived from force laws just like in Newtonian mechanics but f=ma doesn't apply. instead we have f = m (γ3aparallel + γaperpendicular)
 
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