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

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Homework Help Overview

This discussion revolves around concepts in Special Relativity, specifically focusing on the definitions and relationships involving 4-force, 4-momentum, energy, and mass. The original poster presents a homework problem that requires justification of Einstein's mass-energy equivalence and exploration of the relationship between 3-force and 3-acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive relationships involving 4-force and 4-momentum, questioning the meaning of constants in their equations. They also explore the implications of the scaling of components of force and acceleration in Special Relativity.

Discussion Status

Participants are actively engaging with the original poster's attempts, providing insights into the relationships between force and acceleration. Some participants suggest that the differences in scaling factors indicate that the 3-force in Special Relativity does not directly correspond to Newtonian mechanics. There is an ongoing exploration of interpretations and implications without a clear consensus.

Contextual Notes

The discussion includes references to specific equations and concepts from Special Relativity, with participants questioning the assumptions and definitions used in the original poster's approach. There is a noted concern about the validity of using external sources for equations.

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

which 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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