Einstein's Relativity Help

[dream chaser]

Hi Guys,

I really need help with a certain aspect of Einstein's Relativity for Physics. I don't know what object/person should be denoted t and what object/person should be denoted t0(proper time). Could someone please be able to explain it to me.

For example, when a spacecraft is going from Earth to Saturn for example, would an observer on Earth be denoted t or t0 and would an observer on the spacecraft be denoted t or t0? and Why?

Also, according to any person's perspective have a t and t0 and be different from each other. For example, the t and t0 for a person on Earth for the scenario above would be different from the t and t0 for a person on the space craft. Consequently, wouldn't both those two people have a t and t0? Also, why do questions like the scenario above presume one person is denoted t and the other is denoted t0 when both people technically have a t and t0?

It will really help me out if someone could help me with this. It will be very much appreciated. Thanks 

[DrDusk]

Hi Guys,

I really need help with a certain aspect of Einstein's Relativity for Physics. I don't know what object/person should be denoted t and what object/person should be denoted t0(proper time). Could someone please be able to explain it to me.

For example, when a spacecraft is going from Earth to Saturn for example, would an observer on Earth be denoted t or t0 and would an observer on the spacecraft be denoted t or t0? and Why?

Also, according to any person's perspective have a t and t0 and be different from each other. For example, the t and t0 for a person on Earth for the scenario above would be different from the t and t0 for a person on the space craft. Consequently, wouldn't both those two people have a t and t0? Also, why do questions like the scenario above presume one person is denoted t and the other is denoted t0 when both people technically have a t and t0?

It will really help me out if someone could help me with this. It will be very much appreciated. Thanks 

t and t_0 depends on who your measuring with respect to. For example if a question says "this event takes time 'T' for an observer on Earth(the event is on Earth), find the time it takes for someone on a spaceship". t_0 in this case would the time it takes for the observer on Earth and t would be the time on the spaceship, this is because the event is happening at rest relative to Earths reference frame.

 Now if the question says "this event takes time 'T' for an observer on a spaceship(the event happens on the spaceship), find the time it takes for someone on the Earth with respect to someone on the spaceship" t_0 would be the time for someone on the spaceship and t would be that for someone on the Earth.

You are completely right in saying each observer will have their own t and t_0, you decide it depending on whose reference frame the event is happening in. For example if the event happens in Earths reference frame, then t_0 will be the the time measured by someone on Earth. If the event happened on the spaceship, t_0 will be the time measured by someone on the spaceship.

t_0 i.e. proper time is measured by whoever is at rest with respect to the event.

[blyatman]

Moving clocks run slower. Just remember that and you can't go wrong. Once you figure out whose time should be longer, you can then figure out whether you need to divide or multiply by the sqrt term.

[dream chaser]

Thank you DrDusk and blyatman for your responses. It is much appreciated.

How would I go about doing this question then.

A star is 27 light-years away according to earth observers. (Note: A light-year is the distance travelled by light inan earth year.) How long would it take a spaceship travelling at 0.90c to reach the star from earth as measured by observers (a) on earth; (b) on the spaceship?

Thanks

[DrDusk]

Thank you DrDusk and blyatman for your responses. It is much appreciated.

How would I go about doing this question then.

A star is 27 light-years away according to earth observers. (Note: A light-year is the distance travelled by light inan earth year.) How long would it take a spaceship travelling at 0.90c to reach the star from earth as measured by observers (a) on earth; (b) on the spaceship?

Thanks

This is not a question of time dilation, but rather of Length contraction.

Picture yourself standing on the Earth and imagine the distance between the Earth and The Sun as a ruler. When you measure it on the Earth you are stationary relative to this ruler, so the distance you measure is L_0 which is 27 light years. Now imagine your on the spaceship, you are then moving with respect to the ruler, so according to you on the spaceship, the ruler must contract in length giving a length L_v.

The people on the spaceship therefore have to travel a distance given by





Then by using the fact that time = distance/speed you can figure out the time it should take someone on a spaceship as measured by someone on The Spaceship, however for someone on the Earth, remember they are stationary with respect to this 'imaginary' ruler and hence they measure proper distance which in this case is 27 light years.

In general if your confused as to if a question requires time dilation or length contraction remember that time dilation comes into play with you are moving with respect to an event and want to find out the new time. In this case the event itself is the journey, how can you move with respect to the journey? You can however move with respect to the 'imaginary' ruler I mentioned which means it must be a length contraction question.

[dream chaser]

This is not a question of time dilation, but rather of Length contraction.

Picture yourself standing on the Earth and imagine the distance between the Earth and The Sun as a ruler. When you measure it on the Earth you are stationary relative to this ruler, so the distance you measure is L_0 which is 27 light years. Now imagine your on the spaceship, you are then moving with respect to the ruler, so according to you on the spaceship, the ruler must contract in length giving a length L_v.

The people on the spaceship therefore have to travel a distance given by





Then by using the fact that time = distance/speed you can figure out the time it should take someone on a spaceship as measured by someone on The Spaceship, however for someone on the Earth, remember they are stationary with respect to this 'imaginary' ruler and hence they measure proper distance which in this case is 27 light years.

In general if your confused as to if a question requires time dilation or length contraction remember that time dilation comes into play with you are moving with respect to an event and want to find out the new time. In this case the event itself is the journey, how can you move with respect to the journey? You can however move with respect to the 'imaginary' ruler I mentioned which means it must be a length contraction question.

Okay, thanks.

The answer for (a) said that t would be the time measured by observers on Earth. For (b), they specified t0 as the time observed by observers on the spaceship and thus multiplied t by sqrt{1-\dfrac{v^2}{c^2}}[/tex]. How is t and t0 denoted as they are?

Also, in regards to Einstein's relativity, how do you know what the event is and who's reference frame is at rest relative to the event that is happening?

Thanks 

[DrDusk]

Okay, thanks.

The answer for (a) said that t would be the time measured by observers on Earth. For (b), they specified t0 as the time observed by observers on the spaceship and thus multiplied t by sqrt{1-\dfrac{v^2}{c^2}}[/tex]. How is t and t0 denoted as they are?

This method will give you the same answer as mine. Both are just different ways of expressing the same thing. However I encourage you to learn and use the method I gave. This is because it is much more intuitive and you don't have to get into the mess as to what t and t_0 is. I really don't like that method and it's definitely not how I would've done it.

Also, in regards to Einstein's relativity, how do you know what the event is and who's reference frame is at rest relative to the event that is happening?

As for this. I don't know how I can explain it. Think about it like it's literally just what is occurring, and a reference frame is at rest relative to it if your not moving with respect to it.

[dream chaser]

Thanks DrDusk. Much appreciated 

[DrDusk]

Thanks DrDusk. Much appreciated 

No worries, glad I could help =)

These questions are confusing, so I really encourage you to make sense of the method I used. It's so much better and will save you from silly mistakes and confusion.