(a) A particle with mass m moves along a straight line under...

(a) A particle with mass $m$ moves along a straight line under the action of a force $F$ directed along the same line. Evaluate the derivative in Eq. $(37.29)$ to show that the acceleration $a=d v / d t$ of the particle is given by $a=(F / m)\left(1-v^{2} / c^{2}\right)^{3 / 2} .$ (b) Evaluate the derivative in Eq. $(37.29)$ to find the expression for the magnitude of the acceleration in terms of $F, m,$ and $v / c$ if the force is perpendicular to the velocity.

Key Concepts

Application of the Chain Rule

The chain rule in calculus is essential for differentiating composite functions, such as the momentum, which depends on velocity both explicitly and through the Lorentz factor. Correctly applying the chain rule enables one to derive the correct expressions for acceleration in both the parallel and perpendicular force scenarios in relativistic mechanics.

Relativistic Dynamics

Relativistic dynamics adjusts the classical mechanics framework to account for the effects encountered at speeds comparable to the speed of light. In this context, the momentum is given by p = m?v, where ? is the Lorentz factor. The dynamics involve applying Newton’s second law to the relativistic momentum, which leads to modified expressions for acceleration when forces are applied.

Lorentz Factor

The Lorentz factor, defined as ? = 1/?(1 - v²/c²), is a central quantity in special relativity. It modifies time, lengths, and dynamical quantities such as momentum and energy. The differentiation of the Lorentz factor with respect to velocity is key to deriving expressions for relativistic acceleration, especially when force and velocity are either aligned or perpendicular.

Longitudinal Acceleration

When a force is applied in the same direction as the particle's velocity, it is known as longitudinal acceleration. The derivative of the momentum in this case yields an acceleration expression that includes the factor (1 - v²/c²)^(3/2) multiplying F/m. This shows that as the velocity approaches the speed of light, the acceleration for a given force becomes significantly reduced due to relativistic effects.

Transverse Acceleration

For a force applied perpendicular to the particle's velocity, the situation differs because the force does not directly alter the speed but changes the direction of the motion. The resulting expression for acceleration involves the Lorentz factor in a different manner compared to the longitudinal case, reflecting the altered relationship between force and the resulting change in momentum in perpendicular directions.

Transcript

00:01 We first look at the parallel case in which the force is parallel to the velocity.

00:05 So in that case, only the magnitude will change for the velocity.

00:10 So what we do is we take the time derivative of the momentum.

00:17 And so we have the relativistic equation of momentum.

00:21 We take the time derivative of it.

00:24 And we're actually going to be taking the time derivative of a product of two quantities.

00:29 So we can go ahead and factor out the mass, assuming that it's not going to change with time.

00:35 And we're going to take the derivative of v as well as one over the square root of 1 minus v squared over c squared.

00:44 So we have to use the product rule in calculus for derivatives.

00:51 And so you see here we're going to have 1 over the square root of 1 minus v squared over c squared times the time derivative of the speed, the velocity or speed, plus v times the time derivative of 1 divided by the square root of 1 minus v squared over c squared.

01:14 So we do that, and then our next step, our first term is not going to change.

01:22 It's just going to have the same form, in other words.

01:27 And what i do is i go ahead and rewrite 1 over the square root of 1 minus v squared over c squared, as 1 minus v squared over c squared to the minus 1 half.

01:38 Because when you take derivatives, you're going to use numerical forms of the powers.

01:47 So we go ahead and continue, and it gets a little bit complicated.

01:59 We bring down the power negative 1 .5.

02:03 We bring out the derivative of 1 minus v squared over c squared.

02:07 We also have a dv over d t that appears in the second term here, and then we also have 1 minus v square over c squared to the minus 3 half that also results...

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