How Do You Determine the Four-Vector Quantities for a Moving Particle?

[Dakkers]

Homework Statement

A particle of mass m has a position along the x-axis as a function of time given by the equation

u = cgt / (1 + g²t²)^1/2

where g is a constant and c is the speed of light.
(a) Find the 4-velocity of the particle.
(b) Express x and t as a function of the proper time of the particle.
(c) Find the 4-force acting on the particle. Does it ever exceed the speed of light?

Homework Equations

u^μ = γ(c, u)

λ = dt / dτ

The Attempt at a Solution

a) Given the first equation, the four-velocity is simply

u^μ = γ(c, cgt / (1 + g²t²)^1/2

I think.
b) To find the position, we take the integral of dx/dt and find

x = (c/g)(1 + g²t²)^1/2

If we let dt = γdτ, then we can easily see t = γτ.

However, this is a problem, as the particle is accelerating (its second time derivative is not zero) and that means γ must change. And, another problem, is that we cannot sub in

γ = (1 - u²/c²)^-1/2

because then we have a recursive definition.
c) I know that the four-force is simply mass x second derivative of the four position (or mass x derivative of four-velocity), but I am not too sure how to differentiate it. I also know that the four-force is (F⁰, F) but I don't know how to find F⁰ :(
PLEASE BE GENTLE. I am a first-year physics student, and my university decided to put general relativity into a first-year course.

 

[vela]

Homework Statement

A particle of mass m has a position along the x-axis as a function of time given by the equation

u = cgt / (1 + g²t²)^1/2

where g is a constant and c is the speed of light.

You mean velocity, right?

a) Given the first equation, the four-velocity is simply

u^μ = γ(c, cgt / (1 + g²t²)^1/2

I think.

That's right. Since you know the speed u(t), you can write down an expression for $\gamma$ as a function of t and simplify it a bit.

b) To find the position, we take the integral of dx/dt and find

x = (c/g)(1 + g²t²)^1/2

If we let dt = γdτ, then we can easily see t = γτ.

However, this is a problem, as the particle is accelerating (its second time derivative is not zero) and that means γ must change. And, another problem, is that we cannot sub in

γ = (1 - u²/c²)^-1/2

because then we have a recursive definition.

Your expression for x(t) is correct, but your reasoning about how t and τ are related isn't. Integrate $\gamma = \frac{dt}{d\tau}$ where you have expressed $\gamma$ as a function of t. (Hint: use the substitution gt = sinh θ.) Then you can write x as a function of the proper time.

c) I know that the four-force is simply mass x second derivative of the four position (or mass x derivative of four-velocity), but I am not too sure how to differentiate it. I also know that the four-force is (F⁰, F) but I don't know how to find F⁰ :(

The derivatives are with respect to the proper time τ. It might become clear how to do this after you have expressions for t(τ) and x(τ) from the previous parts of the problem.

 

[Dakkers]

Your expression for x(t) is correct, but your reasoning about how t and τ are related isn't. Integrate $\gamma = \frac{dt}{d\tau}$ where you have expressed $\gamma$ as a function of t. (Hint: use the substitution gt = sinh θ.) Then you can write x as a function of the proper time.


The derivatives are with respect to the proper time τ. It might become clear how to do this after you have expressions for t(τ) and x(τ) from the previous parts of the problem.

Our professor never actually mentioned hyperbolic functions :\

 

[Dakkers]

I'm not too sure how to integrate γ, either. I am so bad at this.

 

[vela]

What expression did you get for $\gamma$ in terms of t? You don't have to use hyperbolic functions. You could use a trig substitution, but I think it works out more simply with sinh.

 

[Dakkers]

I got

γ = (1 / (1 + g²t²))^-1/2

 

[vela]

Good. You can simplify it a bit more:
$$\gamma = \sqrt{1+(gt)^2}.$$So now you have
$$\frac{dt}{d\tau} = \sqrt{1+(gt)^2},$$ which you can solve to find $t(\tau)$.

 

[Dakkers]

would it simply be

t = (1 + (gt)²)^1/2τ

?

 

[vela]

No, you can't do that because t varies as τ does. What you have to do is get t on one side and τ on the other:
$$d\tau = \frac{1}{\sqrt{1+(gt)^2}}\,dt$$and then you can integrate both sides to get
$$\tau = \int \frac{1}{\sqrt{1+(gt)^2}}\,dt$$