- #1

- 3

- 0

v(t)=at

and

d(t)=0.5at

^{2}

(assuming x

_{0}=v

_{0}= 0)

Thanks!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter servo75
- Start date

- #1

- 3

- 0

v(t)=at

and

d(t)=0.5at

(assuming x

Thanks!

- #2

Bill_K

Science Advisor

- 4,157

- 201

x = (c

t = (c/a) sinh(aτ/c)

where τ is the particle's proper time. The velocity is v = x/t = c tanh(aτ/c), which is always less than c.

For small τ we can expand the functions and recover the Newtonian formulas:

x ≈ (c

t ≈ (c/a)(aτ/c + ... ) = τ

v ≈ c(aτ/c - ...) = aτ

(Note that these nice formulas displace the origin, and use x

- #3

Bill_K

Science Advisor

- 4,157

- 201

Sorry, should be v = dx/dt = (dx/dτ)/(dt/dτ) = (c sinh(aτ/c))/(cosh(aτ/c)) = c tanh(aτ/c)The velocity is v = x/t = c tanh(aτ/c)

- #4

jtbell

Mentor

- 15,809

- 4,063

See also http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

Last edited by a moderator:

- #5

- 529

- 22

- #6

Bill_K

Science Advisor

- 4,157

- 201

1977ub, I believe you misread the question. We're not talking about the effect of an applied external force, presumably constant in the initial rest frame. This is a spaceship accelerating under its own thrust, constant in its own instantaneous rest frame.

Last edited:

- #7

Meir Achuz

Science Advisor

Homework Helper

Gold Member

- 3,533

- 114

This is a standard problem. The acceleration g is in the instantaneous rest system of the spaceship, not in the original fixed system.

For an acceleration g in the instantaneous rest system of the spaceship, the velocity is (with c=1)

[tex]v=\frac{gt}{\sqrt{1+g^2t^2}}=\frac{dx}{dt}[/tex].

The distance traveled is

[tex]x=\left(\sqrt{1+g^2t^2}-1\right)/g[/tex].

For an acceleration g in the instantaneous rest system of the spaceship, the velocity is (with c=1)

[tex]v=\frac{gt}{\sqrt{1+g^2t^2}}=\frac{dx}{dt}[/tex].

The distance traveled is

[tex]x=\left(\sqrt{1+g^2t^2}-1\right)/g[/tex].

Last edited:

- #8

- 529

- 22

1977ub, I believe you misread the question. We're not talking about the effect of an applied external force, presumably constant in the initial rest frame. This is a spaceship accelerating under its own thrust, constant in its own instantaneous rest frame.

Actually I didn't finish my thought with that post because I couldn't figure out / remember certain things about the issue in the ship's perspective. It's pretty straightforward to decide the velocity of the ship from the perspective of the original frame, and issues which stand in the way of success of getting to c for an external agent.

For someone on the ship, how do they determine what is standing in the way of getting to c? How do they determine their rate wrt the original frame at any given moment?

If they trust / use the original frame's mileposts, which are encountered to be coming closer and closer together, will the decide that they must be > c wrt the original frame?

Last edited:

- #9

- 31,852

- 8,713

If they divide distance in the launch frame by their own proper time then they will indeed get a number larger than c. This is not a violation of SR because the specified quantity is not a velocity.If they trust / use the original frame's mileposts, which are encountered to be coming closer and closer together, will the decide that they must be > c wrt the original frame?

- #10

- 529

- 22

If they divide distance in the launch frame by their own proper time then they will indeed get a number larger than c. This is not a violation of SR because the specified quantity is not a velocity.

Ok Thanks that's what I thought. [everyone on the ship breaks out the champagne]

Share: