Basic special relativity calculation

In summary, the two ships have traveled the same distance after 40 years in the Earth's frame. However, the captain of rocket 1 has aged by 4 years more.
  • #1
marmoset
42
6

Homework Statement



A spaceship sets off from Earth for a distant destination, traveling in a straight line at uniform speed [tex]\frac{3c}{5}[/tex]. Ten years later a second ship sets off in the same direction at [tex]\frac{4c}{5}[/tex].

i) Consider the distance traveled by each spaceship as a function of time to work out when (in the Earth frame) the second spaceship will have traveled as far as the first.

ii) Calculate which of the ship's captains has aged the most during their journey, and by how much.

Homework Equations



The lorentz transformations, and [tex]speed = \frac{distance}{time}[/tex]


The Attempt at a Solution



I have a solution, but it's one of the first SR problems I've done so I am unsure whether it's correct.

Using c = 1 lightyear per year.

i) In Earth frame:

distance traveled by first rocket = distance traveled by second rocket

[tex]\frac{3ct}{5} = \frac{4c(t-10)}{5}[/tex]

[tex]3t = 4t - 40[/tex]

[tex]t = 40[/tex]

So the rockets have traveled the same distance after 40 years in the Earth's frame.


ii) Where x is distance traveled in Earth's frame when rockets meet:

x = speed of rocket 1 * time
[tex] = \frac{3}{5} * 40[/tex]
= 24 light years

time elapsed in rocket 1's frame = [tex]\frac{1}{\sqrt{1 - \frac{3^2}{5^2}}} * (40 - \frac{3*24}{5})[/tex]
[tex]=\frac{5}{4} * 25.6[/tex]
= 32 years

time elapsed in rocket 2's frame = [tex]10 + \frac{1}{\sqrt{1 - \frac{4^2}{5^2}}} * (30 - \frac{4*24}{5})[/tex]
[tex]= 10 + \frac{5}{3} * 10.8[/tex]
= 28 years

So the captain of rocket 1 has aged by 4 years more.

Is this the correct way to approach this problem, and is the answer correct?

Thanks for your help : )
 
Last edited:
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  • #2
Looks good to me.
 
  • #3
Thanks.

Another question: I have seen the lorentz transform for time, which I know as

[tex]t new = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} * (t - \frac{x*v}{c^2})[/tex]

written as

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

I understand how to get to this form by taking a t out of the brackets, but my question is can I apply the new form in all the same places I use the old one? And if so, why does anyone use the first form, as it seems more complicated?
 
  • #4
That second formula is a special case that only applies when the time is that measured by a single moving clock (what you call t_new). It says that when viewed from another frame, a moving clock runs slow--the so-called time dilation formula.

No, those formulas are not interchangeable.
 
  • #5
Ok, but for the first formula the time coordinate in the new frame depends on the x coordinate of the event in the old frame. So if I had two events that happened in the rest frame at different x coordinates but the same t, y, and z coordinates coordinates, then their new t values in the moving frame would be different. Is this correct, and if so is there an intuitive way of understanding it?

edit: OK I think this is the thing about events being simultaneous in some frames but not in others, but I'd still like to understand it intuitively.
 
Last edited:
  • #6
You are exactly right about it having to do with the relativity of simultaneity. You can deduce from the Lorentz transformation:
[tex]\Delta t' = \gamma(\Delta t - v\Delta x/c^2)[/tex]

that two events happening at the same time but different x coordinates in the unprimed ("rest") frame will take place at different times in the primed frame:
[tex]\Delta t' = \gamma(0 - v\Delta x/c^2) = -\gamma(v\Delta x/c^2)[/tex]

If you imagine two synchronized (unprimed) clocks present at these two events you can deduce that they will be unsynchronized according to the primed frame.

In my opinion, there are three ways to understand basic special relativity and all three are important:

(1) Using the Lorentz transformations. For certain problems, this is the best approach, but often it leaves intuition behind.

(2) Using the relativistic behavior of clocks and measuring rods. This can be extremely helpful in developing intuition. The three relativistic behaviors are: Length contraction, time dilation, and--the one you brought up--clock desynchronization (or the relativity of simultaneity). Some problems can be solved much quicker if you have a solid understanding of these relativistic effects.

(3) Using space-time diagrams. This is where you begin to develop a real understanding of relativity.

You might find this summary helpful: Basic Equations of Special Relativity
 
  • #7
That is really useful. I have started reading about spacetime diagrams, I think it will take a while for me to understand them though.

Thanks for helping me, it's been really helpful : )
 

1. What is special relativity?

Special relativity is a theory developed by Albert Einstein in 1905 to explain the relationship between space and time. It describes how the laws of physics are the same for all observers in uniform motion, and how the speed of light is constant in all inertial frames of reference.

2. What is the basic equation used in special relativity calculations?

The basic equation used in special relativity calculations is the Lorentz transformation, which calculates how measurements of space and time change between two frames of reference moving at a constant velocity relative to each other. It is represented by the equations:

x' = (x-vt)/√(1-v²/c²) and t' = (t-vx/c²)/√(1-v²/c²), where x and t are the measurements in one frame of reference, x' and t' are the measurements in the other frame of reference, v is the relative velocity between the two frames, and c is the speed of light.

3. How is time dilation explained in special relativity?

Time dilation is a phenomenon predicted by special relativity where time appears to pass slower for objects in motion compared to those at rest. This is due to the fact that the faster an object moves, the more it experiences a distortion of time and space. The equation for time dilation is t' = t/√(1-v²/c²), where t is the time in the stationary frame and t' is the time in the moving frame.

4. What is length contraction in special relativity?

Length contraction is another consequence of special relativity, where an object in motion appears shorter in the direction of its motion compared to when it is at rest. This is due to the fact that as an object moves faster, its length contracts in the direction of motion. The equation for length contraction is x' = x/√(1-v²/c²), where x is the length in the stationary frame and x' is the length in the moving frame.

5. Can special relativity be applied to everyday situations?

Yes, special relativity has been experimentally proven and is used in many modern technologies, such as GPS systems and particle accelerators. While the effects of special relativity are only noticeable at very high speeds, its principles and equations can still be applied to everyday situations to better understand the relationship between space and time.

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