Time Dilation in Space Travel: Observations from Earth and the Spacecraft

[zebra1707]

Hi there

Can members please review my understanding of the problem.


1. The problem

A star is 522 light years away from Earth. Imagine that the spacecraft was able to travel to the star at a speed pf 0.9999c

a) How long does the trip take, as observed from Earth?
b) How mauch time has passed on the spacecraft s clock, as seen from Earth?
c) How much time has passed on Earths clock, as seen from the spacecraft ?

Homework Equations

Part a) t = d/v 522 light years /0.9999c = 522.05 years (I think that this okay)


Part b) to = 522.05 √ 1 – (0.9999c)^2
=7.38 years
=7 years 139 days (not sure if I need to convert to years and days or leave in years?)


Part c) This is the part that I am difficulty with? I think that the time on Earth, as observed by the spacecraft would be as follows:

tv = 7.38/ √ 1 – (0.9999c)^2
=??

The Attempt at a Solution

As above

 

[Andrew Mason]

Think symmetry.

In b) you are trying to find the time coordinates of an event as measured in the spacecraft frame knowing what the time co-ordinates in the Earth frame are. Using the Lorentz transformation:

(1) $$t' = \gamma(t - vx/c^2)$$

where t' is the time measured in the Earth frame and t and x are measured in the spacecraft frame.

The inverse Lorentz transformation transform Earth time to time in the spacecraft frame:

(2) $$t = \gamma(t' - vx'/c^2)$$

where t is the time measured in the spacecraft frame and t' and x' are measured in the Earth frame.In this case, the two events are:

1. spacecraft passes Earth at (x1,t1) = (x1',t1') = (0,0) in both frames

2. spacecraft passes star at (x2, t2) in its frame and at (x2', t2') as measured in the Earth frame.

Since both events happen at x = 0 in the spacecraft frame, the second event co-ordinates in the spacecraft frame are:

$$(x_2, t_2) = (0, t_2)$$

The second event co-ordinates in the Earth frame are:

$$(x_2', t_2') = (522cy, x'/v) = (522cy, 522cy/.9999c)$$Plugging these values into (1) above:

$$t_2' = \gamma(t_2 - vx_2/c^2) = \gamma(t2)$$

$$t_2 = t_2'/\gamma = 522.05/70.71 = 7.38 y$$

That is what you have done, but I have done it the long way.

For part c) you want to calculate the time of the second event as seen on the Earth's clock in the spacecraft frame. This is a bit different since that event is 522 light years away from the earth. But you will get it right if you stick to the inverse Lorentz transformation:

(2) $$t_2 = \gamma(t_2' - vx_2'/c^2)$$

where x2' = 522cy, v = .9999c and t2' = 522cy/.9999c Work that out.

AM

 

[zebra1707]

Hi AM

Our physics course does not use gamma - hence I am a little confused.

Can you clarify this part of the equation?

Cheers P

Think symmetry.

In b) you are trying to find the time coordinates of an event in the spacecraft frame as observed in the Earth frame. Using the Lorentz transformation:

(1) $$t' = \gamma(t - vx/c^2)$$

where t' is the time measured in the Earth frame and t and x are measured in the spacecraft frame.

The inverse Lorentz transformation transform Earth time to time in the spacecraft frame:

(2) $$t = \gamma(t' - vx'/c^2)$$

where t is the time measured in the spacecraft frame and t' and x' are measured in the Earth frame.


In this case, the two events are:

1. spacecraft passes Earth at (x1,t1) = (x1',t1') = (0,0) in both frames

2. spacecraft passes star at (x2, t2) in its frame and at (x2', t2') as measured in the Earth frame.

Since both events happen at x = 0 in the spacecraft frame, the second event co-ordinates in the spacecraft frame are:

$$(x_2, t_2) = (0, t_2)$$

The second event co-ordinates in the Earth frame are:

$$(x_2', t_2') = (522cy, x'/v) = (522cy, 522cy/.9999c)$$


Plugging these values into (1) above:

$$t_2' = \gamma(t_2 - vx_2/c^2) = \gamma(t2)$$

$$t_2 = t_2'/\gamma = 522.05/70.71 = 7.38 y$$

That is what you have done, but I have done it the long way.

For part c) you want to calculate the time of the second event as seen on the Earth's clock in the spacecraft frame. This is a bit different since that event is 522 light years away from the earth. But you will get it right if you stick to the inverse Lorentz transformation:

(2) $$t_2 = \gamma(t_2' - vx_2'/c^2)$$

where x2' = 522cy, v = .9999c and t2' = 522cy/.9999c


Work that out.

AM

 

[diazona]

Gamma is just an abbreviation.
$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

 

[zebra1707]

Hi there

I calculated 4.10 x 10^-13 years, but this sound totally wrong. Guidence appreciated.

Cheers P

 

[resaypi]

why should part b and part c have different answers, can you briefly explain that. what I think about it is since reference frames are the same, they should have the same answer. and using the metric $$T^2 = t^2 - x^2 / c^2$$ where $$T$$ is the time for the Earth frame and $$t$$ is the time in Earth frame and
$$x$$ is the length in Earth frame. i find the answers for part b and c to be the same

 

[Andrew Mason]

why should part b and part c have different answers, can you briefly explain that. what I think about it is since reference frames are the same, they should have the same answer. and using the metric $$T^2 = t^2 - x^2 / c^2$$ where $$T$$ is the time for the Earth frame and $$t$$ is the time in Earth frame and
$$x$$ is the length in Earth frame. i find the answers for part b and c to be the same

You are right. By symmetry they should be the same since SR postulates the equivalence of all inertial frames of reference. I am not sure why you think they have different answers. Have you worked out t2 ?(the Earth co-ordinates of the event translated to the spacecraft frame):

(2) $$t_2 = \gamma(t_2' - vx_2'/c^2)$$

where x2' = 522cy, v = .9999c and t2' = 522cy/.9999c ?

AM

 

[Andrew Mason]

Hi there

I calculated 4.10 x 10^-13 years, but this sound totally wrong. Guidence appreciated.

Cheers P

Work it out again.

$$t_2 = \gamma(t_2' - vx_2'/c^2)$$ where

$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = ?$$

$$t_2 = \gamma(522cy/.9999c - .9999c * 522cy/c^2)$$

(hint: all the c's cancel out so it is quite easy - answer in years - y)

AM

 

[zebra1707]

Hi there

using gamma = (0.9999c = 70.7) eg (for 0.999c = 22.37)

I end up with a calculation of 36,908.93 years?

It seems a huge number of years.

Cheers P

 

[Andrew Mason]

Hi there

using gamma = (0.9999c = 70.7) eg (for 0.999c = 22.37)

I end up with a calculation of 36,908.93 years?

It seems a huge number of years.

Cheers P

You are simply multiplying 522.05y by 70.71. Why are you doing that? You have to use the Lorentz transformation that I set out for you:

$$t_2 = \gamma(t_2' - vx_2'/c^2)$$

where t2' = 522cy/.9999c, v = .9999c and x2 = 522cy

AM

 

[zebra1707]

Hi AM

t2 = 70.7 * (522.05 - (0.9999)(522)/(3.00^8)2 still get a strange figure.

t2 = 8.02^-17 again this looks strange. Sorry to be such a pain.

 

[resaypi]

Why don't you use the metric, it is much simpler.
$$dT^2 = dt^2 - dx^2 / c^2$$ and $$dT^2 / dt^2 = 1 - v^2 / c^2$$

 

[zebra1707]

Hi resaypi - this has confused me, sorry.

 

[Andrew Mason]

Hi AM

t2 = 70.7 * (522.05 - (0.9999)(522)/(3.00^8)2 still get a strange figure.

t2 = 8.02^-17 again this looks strange. Sorry to be such a pain.

There is no need to actually use the speed of light. The c's cancel out.

$$t_2 = \gamma(t_2' - vx_2'/c^2)$$

t2' = 522cy/.9999c = 522y/.9999
v = .9999c
x2' = 522cy

So:

$$t_2 = \gamma(522cy/.9999c - .9999c*522cy/c^2) = \gamma(522.0522 - 521.9478)y$$

AM

 

[resaypi]

$$T$$ is the time for the Earth frame $$t$$ is the time in Earth frame and
$$x$$ is the length in Earth frame

$$T^2 = t^2 - x^2 /c^2$$ is a direct consequence of the lorentz transformation. Sorry for the lack of explanation before. Does this help better now?

 

[Andrew Mason]

$$T$$ is the time for the Earth frame $$t$$ is the time in Earth frame and
$$x$$ is the length in Earth frame

$$T^2 = t^2 - x^2 /c^2$$ is a direct consequence of the lorentz transformation. Sorry for the lack of explanation before. Does this help better now?

T is the proper time of the event - that is, it is the time between events that occur at the same location in a frame of reference.

In this case the events ( spacecraft leaving Earth and then arriving at star) occur at different spatial co-ordinates in the Earth frame (ie. (x',t') = (0,0) and (522cy, 522cy/.9999c)). In the frame of reference of the spacecraft the spatial co-ordinate is the same for both events (0). The co-ordinates are $(0,0) and (0,\tau)$ where $\tau$ is the proper time:

$$\tau^2 = (522cy/.9999c)^2 - 522cy^2/c^2) = 54.50y^2 = (7.38y)^2$$

This is the same as applying the Lorentz transformation:

$$t' = \gamma (t - vx/c^2)$$ (where t' is the Earth time co-ordinate and (x,t) are the space time co-ordinates of the event in the spacecraft frame.

between the two frames (that is translating the spacecraft time and space co-ordinates to Earth time) as I have shown (see first part of post #2).

Translation of the spacecraft co-ordinates to the Earth frame gives you how the spacecraft 's measurement of the space and time co-ordinates of the event ( spacecraft passing star) is viewed by the earth.

In part c), you are trying to do the inverse: translate the coordinates of an event in the Earth frame to the spacecraft frame. Use the Lorentz transformation:

$$t = \gamma(t' - vx'/c^2)$$

where t' and x' are the co-ordinates of the event as measured in the Earth frame. t, then, is the translation of the Earth time co-ordinate of the event to the spacecraft frame. This is how the spacecraft views Earth's measurements of the space and time co-ordinates of the event ( spacecraft passing star).

It works out to the same as the proper time because of the symmetry (if there was a difference, the two inertial frames would not be equivalent). So you can use the argument that it has to be the same, but that does not show the physics. The physics requires use of the Lorentz transformation.

AM

 

[resaypi]

So you can use the argument that it has to be the same, but that does not show the physics. The physics requires use of the Lorentz transformation.

Why?? The metric is much more natural than the Lorentz transformation, it describes the geometry of spacetime more effectively and intuitively.
Solving equations is not physics, like solivng the lorentz transformation, and in this case the metric provides a much better comprehension of relativity.

 

[Andrew Mason]

Why?? The metric is much more natural than the Lorentz transformation, it describes the geometry of spacetime more effectively and intuitively.
Solving equations is not physics, like solivng the lorentz transformation, and in this case the metric provides a much better comprehension of relativity.

The metric

$$s^2 = (\Delta t')^2 - (\Delta x')^2/c^2 = (\Delta t)^2 - (\Delta x)^2/c^2$$

relating to two events, is invariant for all inertial frames of reference.

So in the event that one of the two frames was the frame in which both events occurred at the same place and if you knew the time and space co-ordinates in the other frame, you could use the space-time metric to determine the proper time. In this case, the proper time is that of the spacecraft frame.

Although you can use the metric for that calculation, which is part b, you cannot use the metric to go the other way. If you plug in the Earth x co-ordinate of the event (522cy) you end up with the time co-ordinate of 522y/.9999. That is just the time measured by the Earth observer as it appears on earth. What you want is the time co-ordinate measured by the Earth observer as it appears to the spacecraft observer.

As I see it, you have to use the Lorentz transformation to translate the Earth frame to the spaceship frame. You can use the metric and the symmetry to predict what it must be, but you need the Lorentz transformation to actually work it out. If you disagree, perhaps you could show us how the metric can be used to determine the co-ordinates of the events as measured by the Earth as observed by the spacecraft observer.

AM

 

[zebra1707]

Hi AM

Thank you both for your asistance - depending on the number of figures used - you could end up with 7.07 rounding or 7.38 years (7.38) which is the same as part b)

(c) How much time has passed on Earth’s clock, as seen from the spacecraft ?

How do you tell which is the most accurate? Cheers P

There is no need to actually use the speed of light. The c's
cancel out.

$$t_2 = \gamma(t_2' - vx_2'/c^2)$$

t2' = 522cy/.9999c = 522y/.9999
v = .9999c
x2' = 522cy

So:

$$t_2 = \gamma(522cy/.9999c - .9999c*522cy/c^2) = \gamma(522.0522 - 521.9478)y$$

AM

 

[Andrew Mason]

Hi AM

Thank you both for your asistance - depending on the number of figures used - you could end up with 7.07 rounding or 7.38 years (7.38) which is the same as part b)

(c) How much time has passed on Earth’s clock, as seen from the spacecraft ?

How do you tell which is the most accurate? Cheers P

Be careful about rounding too early. What you are trying to calculate is the difference between xy and x/y where y is very close to 1. In order to do this, take out the x factor and calculate 1/y - y

$$t_2 = \gamma(522(\frac{1}{.9999} - .9999) = \gamma (522(1.0001 - .9999) = \gamma*.1044$$

AM

 

[zebra1707]

Be careful about rounding too early. What you are trying to calculate is the difference between xy and x/y where y is very close to 1. In order to do this, take out the x factor and calculate 1/y - y

$$t_2 = \gamma(522(\frac{1}{.9999} - .9999) = \gamma (522(1.0001 - .9999) = \gamma*.0144$$

AM

Hi there AM - many thanks for your patience - its not an easy subject to get my around.

Should that be $$\gamma$$*0.1044?

Cheers P

 

[Andrew Mason]

Hi there AM - many thanks for your patience - its not an easy subject to get my around.

Should that be $$\gamma$$*0.1044?

Cheers P

Yes. .1044. It works out to 7.38 years. And I think you meant 'mind'.

AM

 

[zebra1707]

Thanks AM - you have the patience of a saint - yes, I did mean "mind".

Many thanks. P

Yes. .1044. It works out to 7.38 years. And I think you meant 'mind'.

AM