Relativistic time, distance and velocity

  • #1
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A ship leaves Earth bound for a star 4.5 light-years away. According to the ship's clocks, the trip took 4.5 years. What was the velocity of the ship in the reference frame of the Earth?

I'm probably wrong, but I say v=c.

In the reference frame of the ship, the star is 4.5 light-years away and it is approaching you at some velocity v. If it takes 4.5 light-years to get to you, then it is approaching you at a velocity of c. The clocks outside the reference frame of the ship are going slower. In fact, they are stopped when v=c.

Using the Lorentz equations I alway have one more unknown than I have equations.
 

Answers and Replies

  • #2
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In the reference frame of the ship, the star is 4.5 light-years away

Is it? Think in terms of the distance between the Earth and the star as measured from the ship's frame.
 
  • #3
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Is it? Think in terms of the distance between the Earth and the star as measured from the ship's frame.

If it isn't, then that just introduces the length contraction equation with three more variables, two of which are unknown.

If you are on the star, then from your reference from the ship is 4.5 light-years away coming towards you at some speed v.

If you are on the ship, then the star is 4.5 light-years away coming towards you at some speed v.
 
  • #4
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If it isn't, then that just introduces the length contraction equation with three more variables, two of which are unknown.
Could you be more explicit? I'm not sure which variables you're referring to.

If you are on the ship, then the star is 4.5 light-years away coming towards you at some speed v.
That's exactly what's NOT happening. As I stated in my earlier post, the distance between the Earth and the star is not 4.5ly in the ship's frame. The star is travelling a shorter distance while moving towards the ship, as per the measurements made in the ship's frame, in a timespan of 4.5 years.
 
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  • #5
Doc Al
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A ship leaves Earth bound for a star 4.5 light-years away. According to the ship's clocks, the trip took 4.5 years. What was the velocity of the ship in the reference frame of the Earth?
Hint: From the Earth frame the ship's clock is a moving clock. All you need is the time dilation formula for moving clocks and the definition of velocity.
 
  • #6
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Could you be more explicit? I'm not sure which variables you're referring to.

Length contraction.

YOu have x', x and v. You know x (4.5 light years), but not x' or v.
 
  • #7
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Hint: From the Earth frame the ship's clock is a moving clock. All you need is the time dilation formula for moving clocks and the definition of velocity.

Which results in a messy equation. The was a question on the GRE physics exam. It's suppose to be somewhat simple.

[tex]t'=\frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

[tex]v=\frac{dx}{dt} = \frac{x}{t}[/tex]

Combining them:

[tex]t'=\frac{t-\frac{x^2}{tc^2}}{\sqrt{1-\frac{x^2}{t^2c^2}}}[/tex]

Solving that for t would be too time consuming for a GRE question.
 
  • #8
robphy
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Can you draw a spacetime diagram of the situation from the earth's frame?
If you think in terms of intervals, you don't explicitly need Lorentz transformations or time-dilation formulas.
If you can do [hyperbolic] trigonometry, the solution is very simple.
 
  • #9
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Which results in a messy equation. The was a question on the GRE physics exam. It's suppose to be somewhat simple.
It is simple. You can solve it using time dilation or length contraction.

Use the formula for time dilation of a moving clock, not the Lorentz Transformation for time.
 
  • #10
Doc Al
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Length contraction.

YOu have x', x and v. You know x (4.5 light years), but not x' or v.
To use length contraction, express the distance as seen by the ship in terms of the Earth frame distance. That's leaves only one variable. (Hint: Start by writing the simplest expression for the velocity.)
 
  • #11
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It is simple. You can solve it using time dilation or length contraction.

Use the formula for time dilation of a moving clock, not the Lorentz Transformation for time.

It still used the Lorentz factor

[tex]\Delta{t}=\frac{\Delta{t_0}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

I get a mess that looks like this:

[tex]v^4-c^2v^2+c^2\sqrt{\frac{\Delta{x}}{\Delta{t}}}[/tex]

I can solve that quadratically for v.
 
  • #12
robphy
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Draw a right triangle.... where you are given a hypotenuse and the [opposite] side. Determine the [adjacent] side... using the interval. Compute the ratio of the (opposite side)/(adjacent side).

[FYI.. but not needed: the adjacent side is related to the time-dilation formula]
 
  • #13
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Now the problem is this:

[tex]a=1[/tex]
[tex]b=-c^2=-1[/tex]
[tex]c=c^2\sqrt{\frac{\Delta{x}}{\Delta{t}}}=\sqrt{\frac{4.5}{4.5}}=1[/tex]

Now when you put that into a quadratic forumla:

[tex]\sqrt{b^2-4ac}=\sqrt{-4}[/tex]
 
  • #14
Doc Al
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It still used the Lorentz factor

[tex]\Delta{t}=\frac{\Delta{t_0}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
That's fine.

I get a mess that looks like this:

[tex]v^4-c^2v^2+c^2\sqrt{\frac{\Delta{x}}{\Delta{t}}}[/tex]

I can solve that quadratically for v.
Just write the velocity as distance over time, as measured by the Earth. You'll get an easy equation for v.
 
  • #15
Doc Al
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Try this:
[tex]v = \frac{\Delta x}{\Delta t}[/tex]

[itex]\Delta x[/itex] is given; figure out [itex]\Delta t[/itex] from time dilation.
 
  • #16
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Try this:
[tex]v = \frac{\Delta x}{\Delta t}[/tex]

[itex]\Delta x[/itex] is given; figure out [itex]\Delta t[/itex] from time dilation.

I did. Everytime I work it I get the square root of a negative number.
 
  • #17
Doc Al
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I did. Everytime I work it I get the square root of a negative number.
What did you put for [itex]\Delta x[/itex] and [itex]\Delta t[/itex]?
 
  • #18
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Perhaps somebody can show me the error

Distance to star in reference frame of Earth: [tex]\Delta{x_0}=4.5[/tex] light years
Time to reach star in reference frame of ship: [tex]\Delta{t}=4.5[/tex] years
Distance to star in reference frame of ship: [tex]\Delta{x}=x[/tex]
Time to reach star in reference frame of Earth: [tex]\Delta{t_0}=t_0[/tex]
Velocity of ship in reference frame of Earth: [tex]v=\frac{\Delta{x_0}}{\Delta{t_0}}=\frac{x_0}{t_0}=\frac{4.5}{t_0}[/tex]
Solving for [tex]t_0[/tex]: [tex]t_0=\frac{4.5}{v}[/tex]

Ship's clock in reference frame of Earth: [tex]\Delta{t}=\frac{\Delta{t_0}}{\sqrt{1-\frac{v^2}{c^2}}}=t=4.5=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

Combining the previous two equations:

[tex]4.5=\frac{\frac{4.5}{v}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

The 4.5s cancel, and rearranging:

[tex]v\sqrt{1-\frac{v^2}{c^2}}=\sqrt{v^2-\frac{v^4}{c^2}}=1[/tex]

Sqiaring both sides:

[tex]v^2-\frac{v^4}{c^2}=1[/tex]

[tex]c=1[/tex] so [tex]c^2=1[/tex] therefore:

[tex]v^2-v^4=1[/tex]

There are no real values for [tex]v[/tex] that can satisfy that equation.

Rewriting it: [tex]v^4-v^2+1=0[/tex]

Using quadratic formula:

[tex]a=1[/tex]
[tex]b=-1[/tex]
[tex]c=1[/tex]

[tex]\sqrt{b^2-4ac}=\sqrt{1-4}=\sqrt{-3}[/tex]

Any questions?
 
  • #19
robphy
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The 4.5s cancel, and rearranging:

[tex]v\sqrt{1-\frac{v^2}{c^2}}=\sqrt{v^2-\frac{v^4}{c^2}}=1[/tex]

This says that [tex] v =\gamma [/tex]... but [tex] v<1[/tex] and [tex] \gamma=\frac{1}{\sqrt{1-v^2/c^2}} \geq 1[/tex]!

I suspect this line:
Ship's clock in reference frame of Earth: [tex]\Delta{t}=\frac{\Delta{t_0}}{\sqrt{1-\frac{v^2}{c^2}}}=t=4.5=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
Can you justify it?

(By the way... this method is unnecessarily complicated... but I'll hold off on my solution (by the method I suggested above) until you get it with your method.)
 
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  • #20
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Can you justify it?

http://en.wikipedia.org/wiki/Time_dilation

[tex]\Delta{t}[/tex] is the time interval between two colocal events (i.e. happening at the same place) for an observer in some inertial frame (e.g. ticks on his clock),
[tex]\Delta{t_0}[/tex] is the time interval between those same events, as measured by another observer, inertially moving with velocity v with respect to the former observer,
[tex]v[/tex] is the relative velocity between the observer and the moving clock,

I see. I got my [tex]t[/tex] and [tex]t_0[/tex] reversed.
 
  • #21
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That, I believe, would result in [tex]v=\frac{1}{\sqrt{2}}[/tex]
 
  • #22
robphy
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"A spacetime diagram is worth a thousand words."

That, I believe, would result in [tex]v=\frac{1}{\sqrt{2}}[/tex]

Great!!
Now, see how much simpler it could be....

"A spacetime diagram is worth a thousand words."


[tex]
\]
\begin{picture}(200,200)(0,0)
\unitlength 2mm
{
\put(1,58){$\perp$}
\put(1,5){$\theta$}

\qbezier(0,50)(0,25)(0,0)\put(0,0){O}
\qbezier(0,0)(0,60)(0,60)\put(0,60){T}
\qbezier(0,0)(20,30)(40,60)\put(40,60){S}
\qbezier(0,60)(20,60)(40,60)
}

{
\qbezier[100](0,0)(30,30)(60,60)
\put(60,60){L}
}
\end{picture}
\[
[/tex]

(OS) invariant interval from O to S: ship proper-time [tex]\Delta s=4.5[/tex]

(TS) earth-frame ship-travel distance (better: spatial separation between events O and S in earth-frame) [tex]\Delta x=4.5[/tex]

(OT) earth-frame ship-travel time (better: temporal separation between events O and S in earth-frame) [tex]\Delta t=?[/tex]

Using intervals and components,
note that [tex]\Delta s^2=\Delta t^2 - \Delta x^2[/tex].
So, [tex]\Delta t^2=\Delta s^2 + \Delta x^2 [/tex] and
thus [tex]v=\frac{\Delta x}{\Delta t}=\frac{\Delta x}{\sqrt{\Delta s^2+\Delta x^2}}=\frac{ (4.5) }{\sqrt{2} (4.5)}=\frac{1}{\sqrt{2}}[/tex]

Using rapidity [tex]\theta[/tex] and hyperbolic trigonometry,
[tex]v=\tanh\theta= \tanh(\sinh^{-1}{\frac{\Delta x}{\Delta s}})=\tanh(\sinh^{-1}(1))=\frac{1}{\sqrt{2}}[/tex]

Using Lorentz Transformations or time-dilation formulas [which are complicated functions of the unknown v],
...too lazy to do... but you can check numerically that [tex]\gamma=\cosh\theta=\cosh(\sinh^{-1}(1))=\sqrt{2}[/tex]
 
  • #23
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Great!!
Now, see how much simpler it could be....

Uhhh, I think Lorentz transforms are simpler, assuming you set them up correctly.
 
  • #24
robphy
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Uhhh, I think Lorentz transforms are simpler, assuming you set them up correctly.


As you wish....
I prefer to draw pictures and to reason geometrically and invariantly...
rather than do algebra where [without at least a clear picture]
one is likely to be merely pushing around some symbols [which, as you see, can easily lead to mixups].

For this problem, I didn't have to deal with any complicated expressions for v.... it sort of just fell out... especially if you can reason trigonometrically.

Lots of things are simple... assuming you set them up correctly.

My $0.02.

EDIT:
As you pointed out earlier... this is a GRE question.
Some will take the long "time consuming" road... involving possibly "messy equations"...
others will avoid the "messy equations" and see that it is really "somewhat simple".
 
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  • #25
Doc Al
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Uhhh, I think Lorentz transforms are simpler, assuming you set them up correctly.
Looks like you got it now:

[tex]v = \frac{\Delta x}{\Delta t}[/tex]

[tex]\Delta x = {4.5 c*years}[/tex]

[tex]\Delta t = (4.5 years)/\sqrt{1-v^2/c^2}[/tex]

So:
[tex]v = \frac{4.5 c*years}{4.5 years}\sqrt{1-v^2/c^2}[/tex]

[tex]v/c = \sqrt{1-v^2/c^2}[/tex]

[tex]v = c/\sqrt{2}[/tex]

While I appreciate robphy's approach of using spacetime diagrams, solving this sort of problem is a breeze if you understand the relativistic behavior of clocks and metersticks. (I do strongly agree that the best way to deepen your understanding of special relativity is to learn how to use spacetime diagrams.) Just plugging into the Lorentz transformations without a clear picture to guide you can often lead to nonsense.
 

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