# The faster you go the sooner you get there - maybe

1. Jan 3, 2014

### mschroth

I was thinking the other day The faster a person goes, the sooner he get to his destination. There has to be a speed at which if he goes any faster, time dilation and length contraction would not get him to his destination any sooner even if he goes faster. How would you calculate that speed?

2. Jan 3, 2014

### ZapperZ

Staff Emeritus
Sorry, say that again?

How does time dilation and length contraction somehow conspire to produce such an effect on his travel time?

Zz.

3. Jan 3, 2014

### ghwellsjr

Maybe if you used the Velocity Addition formula and kept adding more and more speed, you would find the limit to be the speed of light. Is that the kind of thing you are looking for?

4. Jan 3, 2014

### mschroth

Okay, if I want to travel from location A to location B, the faster I go the sooner I get there (lets assume I want to get to be to see a certain event and I need to get there very fast). If I approach the speed of light, time dilation would mean that time is slower for me in my vehicle that the area over which I am traveling. Therefore, when I get to B, more time has passed when I stop and therefore I didn't get to B any faster relative to the area surrounding A to B. There has to be a speed at which going any faster would not get me there any sooner so there has to be a time/distance between two points limit. (if the event at B I want to see occurs at a certain time compared to A - lets say one hour - I will not be able to get there to see that event no matter how fast I go depending on the distance from A to B.)

5. Jan 3, 2014

### ZapperZ

Staff Emeritus
This is incorrect. Your time doesn't change with respect to you, no matter at what speed you are traveling. Your proper time never change with respect to speed.

This is a common misconception of Special Relativity. The time dilation is the time that YOU observe of the time of those people on A and on B. They are the ones, according to you, who are having a slower time, not you! So if anything, those people are seeing you get from A to B in an even shorter time, according to you, then had there not been any relativistic effects.

Zz.

6. Jan 3, 2014

### PAllen

Let's break it down to two questions:

1) I want to travel from Earth to Mars and arrive at Mars in the shortest time as measured by the "International Solar Standard Time". Then the the best you could do is to approach but never reach distance/c .

2) I want to travel from Earth to Mars in the shortest time on my wristwatch. Note, I assume you have acceleration and deceleration phases so you can land. Then the only lower bound on the time needed per your wristwatch is zero. In 'principle', you could make the trip in one microsecond per your watch. However, the time per ISST (see (1)) would not detectably less than if you traveled such that your watch time was a minute.

Last edited: Jan 3, 2014
7. Jan 3, 2014

### mschroth

So the way I see it from the explanations that you people gave me is that you could get there faster the faster you go. If you are at the speed of light, you would be there instantaneously.

8. Jan 3, 2014

### PAllen

Only as experienced by the traveler. If you were going to the nearest star and back, you would find earth had passed over 8 years, even though you experienced only a week (for example) on the round trip. Of course, you can never quite reach the speed of light.

9. Jan 3, 2014

### ViperSRT3g

The way you view time wouldn't change because your point of reference (yourself) hasn't changed. Therefore, time would appear to be passing as it normally would. Time outside of your spaceship would appear to be in fast forward.

From an observer outside of your ship, they would be seeing time for themselves passing by at the usual pace. (If they could see) a clock on your spaceship, it would be ticking at an extremely low rate.

This is due to the effect of apparent time slowing down at faster relative speeds. This effect is so pronounced, that it can be measured at distances of less than a meter.

PS: If I've said anything wrong, please correct me.

10. Jan 3, 2014

### mschroth

The problem lies in events outside "my spaceship". I want to go to the nearest star and then get back in time for my son's next birthday. I can't go fast enough to ever do that. However, if I fly from here to Mars, I might be able to swing it. There has to be a distance/time/speed limit where going faster will not accomplish the mission of getting back in time for my son's birthday.

11. Jan 3, 2014

### ZapperZ

Staff Emeritus
The limit is c. But this has nothing to do with the erroneous idea that time dilation/length contraction put a limit on such a thing. You are using the effect as being the cause.

Zz.

12. Jan 3, 2014

### ViperSRT3g

There is no fine point where distance covered, intersects with speed traveled. The faster you go, the sooner you will arrive at your destination, hands down. And the fastest you can travel is c.

13. Jan 3, 2014

### George Jones

Staff Emeritus
You can't if you don't take your son with you. If you do take your son with you, you can, in principle, be back in time to celebrate here on Earth your son's next birthday.

14. Jan 3, 2014

### mschroth

Thanks for all of your insight. But to make a long story short, if I had a spaceship that could travel at or close to the speed of light, I would probably take whoever wanted to come and explore space "the final frontier" and wouldn't worry to much about the 100's or 1000's of years that go by on earth.:surprised

15. Jan 3, 2014

### ZapperZ

Staff Emeritus
But that isn't a physics issue, which makes this quite off-topic.

It is still isn't clear if you have understood where you got Special Relativity wrong, especially in the application of time dilation and length contraction.

Zz.

16. Jan 4, 2014

### Bill_K

Viper, It's unclear what you mean by the underlined sentence, but I hope you're not saying that an observer on the spaceship would perceive Earth's clocks to be running faster!

Inertial reference frames are equivalent, this is the whole point of relativity. Time dilation is symmetrical, and to an observer on the ship, Earth's clocks would appear to be running slower as well.

Last edited: Jan 4, 2014
17. Jan 4, 2014

### PAllen

To add a bit to Bill_K's point, as in many such situations, it is important to distinguish what is seen (directly measured) from what is modeled by a logical model. If we assume a complete trip from earth to the nearest star, you might (for example) accelerate to .9998c, coast, then decelerate. If you watch a clock (suppose you can) on the destination star, visually you will see it advance 8 years while your clock advances 1 month. Now the question is what times that you see correspond to what times in your past (since, up until the end, they are light delayed). At the start, it is obvious that you are seeing times from 4 years ago due to light delay. Thus, the first 4 yours of time you literally see go by are just delayed reception of light emitted before you left. However, the second 4 years you see go by are more interesting. Since your trip involves brief acceleration, coasting, and brief deceleration, you have to have a model of how to map this second 4 years of history you see to times along your trip. There is no strongly preferred way to do this for such a non-inertial trip. But any reasonable way would involve a period, related to the initial acceleration, where you consider the distant clock to be going fast. Following this, during some or all of your coasting period (whether it is some or all depends on what non-inertial simultaneity convention you chose), you would consider the distant clock to be running slow. Why? Because after accounting for light delay (= non relativistic Doppler), you would find that distant clock clearly running slow. This is what Bill_K is referring to.

18. Jan 4, 2014

### Bill_K

Possibly. But if it is, I don't recognize it.

What I had in mind was a lot simpler! no light delays, no acceleration. Those features are important to explain the twin paradox, where the relativistic effects are asymmetric. But as I said, the present situation is symmetric. Explaining it involves nothing more than the bare bones Lorentz transformation and its inverse.

(1) x' = γ(x - v t)
(2) t' = γ(t - v/c2 x)

(3) x = γ(x' + v t')
(4) t = γ(t' + v/c2 x')

Here (x, t) are the coordinates in Earth's inertial reference frame, while (x', t') are the coordinates in the ship's inertial frame.

The ship is located at x' = 0, so from Eq.(4) we have t = γt'. That is, since γ > 1, from the ship's perspective the ship's time t' is less than the Earth's time t.

On the other hand, the Earth is located at x = 0, so from Eq. (2) we have t' = γt. That is, from the Earth's point of view, the Earth's time t is less than the ship's time t'.

19. Jan 4, 2014

### PAllen

Except here there was clearly the idea of trip starting from earth and ending at some star or planet. You have to account for the fact that people at the destination will tell our traveler that he left 4 years ago, and that the traveler will see 8 years of history at destination go by. The ordinary concept of trip has you starting at rest with respect to your origin, and ending at rest with respect to your destination.

20. Jan 5, 2014

### ghwellsjr

Somehow, I'm having trouble reconciling these two statements. Could you please remove my confusion?